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M50Paul
04-25-2005, 07:07 AM
How do you calculate teh following:
1. How many AK combinations are there when 5 cards are seen none of which are AK.

2. How many AA combinations are there with no cards seen

Thanks

PaultheS
04-25-2005, 09:43 AM
for AK: 4 * 4 = 16 combinations. (ie. Ah can go with any of the four kings, Ac can go with any of the four kings, etc.)

for AA: 4C2 = 4! / 2!2! = 6 combinations.

M50Paul
04-25-2005, 03:40 PM
thanks

LetYouDown
04-27-2005, 04:11 PM
[ QUOTE ]

1. How many AK combinations are there when 5 cards are seen none of which are AK.

[/ QUOTE ]
This doesn't change the number of combinations. Are you looking for the odds of having the hand?

M50Paul
04-27-2005, 10:33 PM
good point, agree. I was making some tables on odds and other measures and I kind of ran two things together. This particular question was getting at an objective way to detemine how to randomly raise 75% of the time. So in the case of AK -- 16 combinations I would raise any time there was no king of clubs in my hand.

LetYouDown
04-27-2005, 10:44 PM
Not to nitpick, but that's not random...that's derived /images/graemlins/grin.gif

M50Paul
04-29-2005, 09:08 AM
Well...... it's a non-random, random secletion of an integral part of my digit which gives the the non bias opportunity to rasie with impunity.

So........